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Let 'u' be the differentiable function of x.

$\frac{\text{d}}{\text{d}x}ln(x) = \frac{1}{x}$

Using chain rule

$\frac{\text{d}}{\text{d}x}ln(u) = \frac{1}{u}\frac{\text{d}u}{\text{d}x}$, u >0

**Prove that : $\frac{\text{d}}{\text{d}x}ln(x) = \frac{1}{x}$
Proof : ** Let f(x) = ln(x), then f(x + h) = ln(x + h)

According to the definition of the derivative,

$\frac{\text{d}}{\text{d}x}f(x) =\lim_{h \rightarrow 0}\frac{f(x + h) - f(x)}{h}$

$\frac{\text{d}}{\text{d}x}ln(x) =\lim_{h \rightarrow 0}\frac{ln(x + h) - ln(x)}{h}$

By using the properties of logarithm,

$\frac{\text{d}}{\text{d}x}f(x)=\lim_{h \rightarrow 0}\frac{\frac{ln(x +h)}{x}}{h}$

= $\lim_{h \rightarrow 0}\frac{ln(1 + \frac{h}{x})}{x.h}$

Multiply by x

=$\lim_{h \rightarrow 0}\frac{ln(1 + \frac{h}{x})}{\frac{h}{x}}.\frac{1}{x}$

= 1.$\frac{1}{x}$ ( since $\lim_{h \rightarrow 0}\frac{ln(1 + x)}{x}=1$)

$\frac{\text{d}}{\text{d}x}ln(x) = \frac{1}{x}$

$\frac{\text{d}u}{\text{d}x}$ = 1

$\frac{\text{d}}{\text{d}u}ln(u) = \frac{1}{u}$

So, $\frac{\text{d}}{\text{d}x}ln(u)$

Using the chain rule,

$\frac{\text{d}}{\text{d}x}ln(u)= \frac{\text{d}}{\text{d}u}ln(u). \frac{\text{d}u}{\text{d}x}$

= $\frac{1}{u}$ .1

= $\frac{1}{u}.\frac{\text{d}u}{\text{d}x}$

$\frac{\text{d}}{\text{d}x}ln(u) = \frac{1}{u}.\frac{\text{d}u}{\text{d}x}$, u >0

(u > 0 since ln of negative number and zero is not possible)

**Example :1 ** Find the derivative of f(x) = $ln(x^{2})$

**Solution : ** f(x) = $ln(x^{2})$

Here we will use a chain rule.

Let u = $x^{2}$

$\frac{\text{d}u}{\text{d}x}$ = 2x

f(x) = ln(u)

$\frac{\text{d}}{\text{d}x}ln(u) = \frac{1}{u}.\frac{\text{d}u}{\text{d}x}$

= $ \frac{1}{u}.2x$

$\frac{\text{d}}{\text{d}x}ln(x^{2}) = \frac{1}{x^{2}}.2x$

$\frac{\text{d}}{\text{d}x}ln(x^{2}) = \frac{2}{x}$

**Example 2 : ** Find the derivative of f(x) = $\frac{1+5x}{ln(x)}$ with respect to x.

**Solution : ** f(x) = $\frac{1+5x}{ln(x)}$

According to the quotient rule,

$\frac{\text{d}}{\text{d}x}[\frac{u}{v}] = \frac{v\frac{\text{d}}{\text{d}x}(u) - u\frac{\text{d}}{\text{d}x}(v)}{[v]^{2}}$

Here u = 1 + 5x and v = ln(x)

$\frac{\text{d}}{\text{d}x}[\frac{1 + 5x}{ln(x)}] = \frac{ln(x)\frac{\text{d}}{\text{d}x}(1 + 5x) - (1 + 5x)\frac{\text{d}}{\text{d}x}(ln(x))}{[ln(x)]^{2}}$

= $\frac{5.ln(x) - (1 + 5x).\frac{1}{x}}{[ln(x)]^{2}}$

$\frac{\text{d}}{\text{d}x}[\frac{1 + 5x}{ln(x)}] = \frac{5.ln(x) -\frac{1}{x}-5 }{[ln(x)]^{2}} $

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